RADAR CROSS SECTION LECTURES by. Distinguished Professor Allen E. Fuhs. Department of Aeronautics. Naval Postgraduate Sc
III
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RADAR CROSS SECTION LECTURES
by DISTINGUISHED
PROFESSOR ALLEN E. FUHS Department of Aeronautics Lui
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NAVAL POSTGRADUATE SCHOOL MONTEREY, CALIFORNIA 93940 0
-
RADAR CROSS SECTION LECTURES
by Distinguished Professor Allen E. Fuhs Department of Aeronautics Naval Postgraduate School Monteezy, CA 93940
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INTRODUCTION
These notes were developed while the author was on Sabbatical at NASA Anes Research Center during FY 1982. The lectures were presented to engineers and scientists at NASA Ames in March-April 1982. In August 1982, the RCS lecturre, aere presented at General Dynamics Fort Worth Division.. T- thoroughly cover the content LECTURE
the following time schedule is HOURS LECTURE TIMEI
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III V
1.50
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required:
LECTURE I.
*
INTRODUCTION TO ELECTROMAGNETIC SCATTERING
1.
Level of Complexity
2.
Features of EM Wave
3.
What Is RCS?
4.
Magnitude of Radar Cross Section
5.
Polarization and Scattering Matrix
6.
Inverse Scattering
7.
Geometrical Versus Radar Cross Section
8. Polarization and RCS for Conducting Cylinder 9.
Far Field vs Near Field
10.
Influence of Diffraction on EM Waves
11.
Relation of Gain to RCS
12.
Antenna Geometry and Beam Pattern
13.
Radar Cross Section of a Flat Plate
"14. Wavelength Regions 15.
Rayleigh Region
16.
Optical Region
17.
Mie or Resonance Scattering
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(~A.'EFUHS1 Before discussing RCS, a perspective is given on the conpexity of problems to be encountered. for numerical solution.
A measure of complexity is the tool required
The tools span from slide rule to CRAY computer.
Since these lectures are prepared mainly for the aerodynamicist, typical aerodynamics problems are given along with classes of RCS problems. The lectures provide sufficient information which allows back-of-envelope calculations in the "Southwest" corner of the graph.
The lectures discuss
in a descriptive way the scientific problems in the "Northeast" corner.
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FEATURES OF EN WAVE 0
Wavelength
A " c/f
c - speed of EM wave-
3E8 =/sec
f - frequency, Hz "0 Electric and Magnetic Fields* -
orientation related to antenna (source)
-•.=zZ ;
z
(/)1/2 on
-E-ohms 0
Polarization -
orientation of the electric vector E-9
-
polarization may be important in determining magnitude of RCS
0 Energy and Power* energy density - energy/volume
-
flux of energy - power/area
-
power
-
-
1
i(eE
22
x
+ pH2 ) = Joules/m3 -
Watts/m2
(amplitude squared)
O Interference field vectors add vectorially; may cause cancellation of waves
*J. C. Slater, Microwave Transmission, Dover, 1959.
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A.E. ; FUHS What is RCS? 0
The RCS of any reflector may be thouht of as the projected area of equivalent isotropic (same In all directions) reflector.
The equivalent
reflector returns the same power per unit solid angle.
0 ECS is an area.
*
0
Meaning of RCS can be seen by arranging a in form:
-s
"'-1 "
2
al, - power intercepted and scattered by target, Watts
*
a•I/
4v
- power scattered in 41T steradians solid angle, Watts/steradian
I rA - power into receiver of area Ar. Watts a - Ar/R
2
-
solid angle of receiver "
seen fro
target, steradian
I A i/1- power reflected to receiver per unit solid angle, Watts/steradian
r r
I rA r\'
( /(A 2
power reflected to receiver per unit solid angle, Watts/steradian
-
0 Meaning of limit R is distance from target to radar receiver. H,* UI' and I1 are fixed. E and Hr vary as t/R in far field. I varies as r
I/R' in far field.
Hece, 0 has a limit as R * S
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Magnitude of Radar Cross Section 0
RCS can be expressed in terms of area.
0
Since RCS Is an area, you can check your formulas for RC{S for din"
'on;
formulas should always have dimensions of Zenith squared.
0
The square meter is usually used as a reference to vxress a as a relative value using decibels. Given oa 28 db 2•
/10
in
?
631 a2
3
o - 0.34 m2, what is o in db
a(db 0
what is 28db 10,j
2 Given
An exmple of calculation
3)
- 10 log 1 0 (0.34) -
an
-
? 4.7 dba
Some typical values are shown for various objects.
Also the mSagnitude
of creeping waves or traveling waves from an aircraft is shcAw. When the RCS due to direct reflection is reduced, RCS from other wave scattering phenomena may become important.
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In this case, the motation and
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6
INVERSE SCATTERIG 0 To quote from P%-ofessor Xennaugh* on the subject of inverse scattering: "One measure of electromagnetic scattering properties of an object is the radar cross section (RC.S) or apparent size, In the early days of radar, it was found that rapid variation of RCS with aspect, radar polarization, and frequency complicate the relation between tk-ue and apparent sizes. As measurement capabilities Improved,. investigations of the variation of RCS with these parameters provided the radar analyst with a plethora of data, but few insights into this relation. In the present context, such data are essential in determining the physical features of a distant target, rather than
an annoying radar anomaly." 0
Inverse scattering provides a nonimaging method to determine t~arget size, shape, etc.
-
0 By appropriately processing the backscattered waveforms or target signature observed in radar receivers, different target shapes may be discriminated and
classified. 0
St.ýalth implies denial of detection; an expanded concept for stealth implies control of backscattered waveform, thereby denying information about target size, shape,
etc.
*Edward M. Kennaugh, "Opening Remarks, Special Issue on Inverse Methods in Electromagnetics," IEEE Transactions on Antennas and Propagation, Vol. AP-29, March, 1981.
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A. E# FUH5 GEOMETRICAL V4RSUS RADAR CROSS SECTION 0
Sphere.
The two areas are drawn to scale.
For a sphere,
a - 'ira 2 frdependent of wavelength in optical region.
Solve for a:
a - SQR(a/7r) - 0.56 meter '
0
Square Flat Plate.
"to X
-
Consider a frequency of 8.5 GHz which corresponds
0.035 a - 3.5 cm.
The cross section for a flat plate is
2 2
4AjTA2 4t[(0.lm) 1
7 0
Aircraft Broadside. to the wave vector k.
2
(0.035)2
The aircraft may have a panel which is normal A large RCS results due to reflection from the
panel.
0
Low RCS Aircraft Broadside.
By a combination of RCS reduction methods, the
aircraft has a smaller RCS than projected area.
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"A*** POLARIZATION AMD
FOR CONDUCTING CYLINDER FCS
0
When A is smaller than a, polarization Is not important for magnitude of RCS.
0
The three regions based on relative size of X compared to a are shown. In both Payleigh and optical regions, the RCS varies smoothly with clanging X. In the Mie region, also known as resonance region, the RCS varies rapidly with changing 1. In optical region, 04
and a, converge to kaL 2
0
Cylinders with small ka are used for radar chaff.
0
A cylinder can be used as a model for estimating RCS of the leading edge of a wing or rudder.
0
Mie region occurs where circumference of cylinder, i.e., 2ira, is nearly equal to wavelength, A.
0
The values of ka for which cylinder diameter, d, equals A and for which cylinder radius, a, equals A are shown in the graph.
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A. E; FUHS
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FAR FIELD vs NEAR FIELD "O The symbol f refers to a fraction of a wavelength.
In far field, the
variation in phase is small over a distance L. 0 In far field, the incident wave can be considered to be a plane wave. 0 The radiation from a dipole illustrates far field and near field for microwaves.
Nk3
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Ee - -- exp[j (.t
-
kr)] sin e [ -
+
I
NEAR FE LD
(k) (kr) Ee -
ep[j(wt -kr)] sin8eL O47r~
FAR ir
FIELD
- dipole moment
,.
- electric inductive capacity
S..
k - 2wr/A w - 2wf (f is frequency here) e
polar angle in polar coordinates
j-
square root of minus one
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k
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"0 The reason antennas are important to RCS ,s that the circular antenn, is "equivalent to a circular disc.
The circular antenna is the source for a
plane wave from an aperture in the form of a circle.
plane wave reflected from a circular disc.
The result is a plane wave
from an aperture (the disc) in the form of a circle. area is equivalent to an antenna. * -reflected
6il
Hence, the reflecting
Antennas have side lobes.
The radiation
by a flat plate or a disc has the same side lobes as an antenna of same shape.
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Consider an incident
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13 RADAR CROSS SECTION OF A FLAT PLATE
-
0 The RCS of a flat plate is obtained from (3
GA
where A equalE the plate area, A
'
p 0 Three different geometries leading to a series of wavefronts moving to the right are illustrated.
The beam in the far field is identical for the
three cases illustrated. 0
The shape of the flat plate does not influence the value of a so long as the smallest dimension of the plate is much longer than a wavelength.
0 A test for whether or not the formula applies is accomplished by comparing X and the square root of A . The result must be "p p
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A: EFUHS 14 WAVELENGTH REGIONS 0
Recall k - 2wr/X; consequently
- Ja 2ira a is a characteristic dimension of the body. 0
A complex shape such as an aircraft may have components spanning all three regions.
For example, the wing leading edge may be in the optical region
while a gun muzzle may be in the Rayleigh region. -
0
The region where ka
1
is known as Mie region or as the resonance region.
Resonance may occur between creeping waves and the specular reflected waves. The numerous wiggles characteristic of the resonance region-may be due to the resonance.
0 The resonance region is difficult to analyze.
In the Rayleigh region, series
expansions using ka as an expansion parameter can be accomplished. optical region, the. expansion parameter is 1/ka. technique is not useful for the Mie region.
The series expansion
For Rayleigh scattering,
the leading term in an expansion may be the electrostatic field.
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15
RAYLEIGH REGION
0 An oblate ellipsoid of revolution is shown in the figure.
When the
distance in the axial direction is small, the oblate ellipsoid models a disc like a penny. -
0
For this case, F is not near unity.
A prolate ellipsoid of revolution is shown in the vievgraph. distance along the axis is emphasized, a wire can be modelled.
When the In that case,
F is not near unity.
""
0
For smooth bodies which do not deviate too much from a sphere, the RCS is independent of polarization or aspect angle.
0
The formula for a can be tested for a sphere. a/na2 Assume F " 1.0.
0.1 when ka
For a sphere
0.33
3
Then, since V - 41a /3 2
44 0~ 32 (Ta) a
64 -
Inserting the value for ka, one finds -/•a2 - 0.084 which is close to the ac:urate value.
6
4 2 (ka) va
33
44
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16
OPTICAL REGION 0
One can apply the formula a
i fplP2 to a sphere.
In that case P1 = p 2
a.
S~Hence, •'.
0
-
ira22
which is the anticipated result. O Optical approximation has the greatest use to calculate specular returns and the associated sidelobes.
"0 Optical approximation may fail when there is a surface singularity such as 2-.
an edge, a shadow, or a discontinuity in slope or curvature.
-
singularities may cause second-order effects which include creeping and
-•
traveling waves.
When specular returns are weak, the RCS may be
dominated by creeping or traveling waves.
0 0.
S'
Surface
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MIE OR RESONANCE SCATTERING To satisfy electrical boundary conditions on a body, a grid with nodes spaced at a small fraction of a wavelength, say X/6, is needed. where A
::
A, E, FUH1• • BIJRZTHROUGH RANGE
0
Symbols have definitions as follows: Ar
radar antenna area, in2
A
i Jamer antenna area, m
Abr
area of radar beam at jammer, m2
-Ab
area of jammer beam at radar, m2 angle of jammer beam, radians
e
angle of radar beam, radians
S
signal at radar due to Jammer, Watts
Sr r
Signal at radar due to reflected power from target which is carrying jammer, Watts
.P
radar power, Watts
"-P
Jammer power, Watts
R
range, meters
x
wavelength, meters
o
RCS of target which is carrying jammer, m
0
Obviously both radar and Jammer must be on same X.
0
Note that S varies as R 2 .
0
Note that S varies as R 4 .
*0
r
A narrow beam for Jammer is not practical since this implies jammer must be aimed.
-
I.
2
Hence A is small.
0
Note that R.b varies as SQR(o).
0
For penetrating aircraft, a small value of Rb is desired.
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WAVELENGTH DEPENDENCE USING FLAT PLATE a o For a flat plate, p1
*
Pp2 2
S
ai
does not change with X.
.p
For a cylinder,
0
For a sphere, p1 - p2 and both are finite and nonzero.
p1 ÷
"
/
'0
Hence A
0
and 2/
i-
•" 2 is
finite and nonzero.
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o
P. O.-
0
The wedge, curved wedge, and cone have at least one zero value for radius
Physical Optics
These can be understood in terms of the formula for a
of curvature.
which is based on an integration of aA/3p along the direction of incident wave motion. 0
In this viewgraph, p is the distance along the direction of incident wave propagation.
0 The equation for a is somewhat analogous to the equation for supersonic potential function.
See page 237 of Liepmann and Roshko.*
* |
*H. W. Liepmann and A. Roshko, Elements oC Gas Dynrmics, Wiley, New York, 1957.
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12
RADAR CROSS SECTION OF A SPHERE
0
The formula valid in Rayleigh region ira 2
-
64-"64 9 (ka)
comes from Lecture 1, Viewgraph 15. 0
In the optical region, co is independent of ka; subscript "o" refers to optical region.
0
In the HIE or RESONANCE region, the cross section is the sum of two contributions.
Electric fields add vectorially; power does not add.
Hence the formula a
A,
v
0
c
applies only at maxima or minima of the curves.
At other locations a phase
angle is required.
0 The meaning of the word "resonance" now becomes apparent.
When the speculkr
and creeping waves have the correct relative phase, one gets "resonance" or an
addition of the two waves. O At point 1, which is a maximum in Mie region, ka is nearly 1.0. -2 * ,ra
1.03
Hence
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ADDITION OF SPECULAR AND CREEPING WAVES 0
One can calculate a for the maximum at point 1 of RCS wave
/lra 2 = 1.03 a0o/a2 ---
= 1.00
/Va /77a2 + /ahra] 2 0
Ira
4.06
In terms of db, precise calculations show that the cross section is 5.7 db hig at point 1.
For the calculations here adb
10 logl0(4.06)
-
-
6.09
which is close. 0
At the minimum at point 2 on the curve ka-
c
1.03(.8)-
1.8 0.237
22
_22
S...2
-
[,~i'---
O"
]2-
0.263
which is close to value of 0.28 0
In summary, the wiggles in the RCS curve in the Mie region are due to
"constructive or destructive interference between specular reflected and I----
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U
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creeping waves.
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L U
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14
DERIVATION OF a -iplP 0
AS
arclength along the reflecting surface angle subtended by AS from center for radius R
AS
r
4.
arclength along the reflected wavefront
.ki
propagation vector for incident wave
ki
propagation vector for reflected wave
R
radius of curvature of the reflected wavefront
p
radius of curvature of reflecting surface
a
solid angle formed by reflected wavefront
Pi
incident power, Watts
-Pr
reflected power, Watts
0 The fact that the angle associated with p, i.e., A81 /2, is one-half of the angle associated with R, i.e., A6,, is an important fact. O In the derivation of the equation, one uses the definition of RCS.
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467 .
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T"r..'F? P,
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As
77 I P, Pr
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A:"E. FUHS l RADAR CROSS SECTION FOR WIRES,
0
RODS,
CYLINDERS AND DISCS
The problem has three characteristics lengths, i.e., a, L, and A, and two ratios, i.e., ka - 21a/X and LI/.
0 The values of L/X and ka determine the RCS. 0 Consider a reference square area which is X on each side.
The various
geometrical figures have the A-square drawn to indicate relative sizes
1
of ka and LIX.
--
O The (L/X)
-
(ka) plane has been divided into three regions.
In the upper
left wiere L/A >> 1 and ka >> 1, the polarization of the wave is not important.
In the corner near the origin where L - a and ka > X and appear to be flat.
p is
in quotes because a surface may
radius of curvature.
(7)
Ordnance and drop tanks contribute to RCS.
(8)
Edge diffraction (like a wedge) occurs at sharp leading edges and trailing edges.
Sharp is p 1.
Blunt edges use a
cylinder as model. (9) (10)
Inlet cavities my give very large RCS. The -udder and elevator may form a right angle dihedral which acts as retrord lec tor.
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3 RELATIVE SIZE OF CONTRIBUTORS TO RCS Missile may have own radar which can have large RCS.
ORDNANCE.
RUDDER-ELEVATOR DIHEDRAL may be big due to action of retroreflector.
*'.
EXHAUST.
Waves can propagate within the cavity and reflect from internal parts.
RUDDERS.
RCS is small except for glint at broadside.
WING.
RCS is small except when viewed so as to see "flat" area.
INLET FOR APU.
The inlets for APU, air conditioning ducts, and gun exhaust
gas ports can be large in certain direction. COCKPIT.
Big contributor to RCS.
GUN MUZZLE. RADOME. FUSELAGE.
Scattering is due to surface discontinuities.
Big anLenna inside acts like a cat's eye in the dark. Recall a -
rP0P2 .
upper or lotr surface.
t"t
141
Usually p1 and P2 are small compared to p of wing
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DIV,
4 AIRCRAFT AT VISIBLE AND MICR0WAVE FREQUENCIES 0
The E( boundary conditions and the wave equations are shown. for free space is
+ 0
I
For propagation inside dielectrics, change k 0 to k, C Three major cavities are illustrated: aircraft radar cavity engine inlet cavity cockpit cavity
The equation
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A. E.FUHS
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13,C rdie leety k 2+
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1
FIRE FOX MIG-31 0 As an example of estimating RCS for an aircraft, the MIG-31 Fire Fox will be used.
0 Some of the gross features of the RCS for the aircraft can be obtained *:)
from formulas discussed earlier.
O The Fire Fox was designed, not for Mach 6, but for movie audiences. Low RCS and good L/D were not requirements.
The design requirement was
to look "tmean."
Note:
The comments for vievgraph 6 start here.
PLAN FORM FIRE FOX MIG-31 The assumed plan form is shown.
The plan form can be verified now that the
MIG-31 is in U. S. hands. Assume the following: length, 28 m wing span, 30 m
radar frequency, 12 GHz
The RCS will bi estimated in the plane of the plan form.
Wavelength, 0.025 u
One views the aircraft from
nose-on (6 - 0 ) moving clockwise to starboard wing tip (8 - 900). SU scattering components are identified.
The various
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6
PLAN FORM FIRE FOX MIG-31 (Continued from Previous Page)
NOSE-ON
e
0=
Tip Diffraction: The tip of the fuselage appears to be a wedge. Formulas for finite wedges are quite complex and are outside the realm of a back-of-anenvelope calculation. Based on calculations for tip diffraction for cones, a. may not be too important. However, this should be verified. Engine Inlets: There are four engine inlets which will be modelled as flat plates with size 0.7 m x 1.8 m for each. -- .
a1I -4irA 22
41(0.7 x 1.8)2 ... (.025)2(.025)2
1
2 32000 m2 - 45 db
One can estimate the width of the main lobe from
= [sin(ka sin D- 2
a
a 0.
kas in e>1
The first zero occurs when ka sin 6 - n.
"e - arcsin(ir/ka) -
•angles
1.020 1
The lobe is very narrow. To add the four inlets, use the equation from Lecture 2, viewgraph 29.
If phase
for the waves returned from the four inlets are all zero, then the addition formula becomes
-
4=
2
16a
The RCS for all four inlets is 57 db. .-LEADING
EDGE OF STARBOARD CANARD
The LE has a sweep of 2 0'. The LE can be modelled as a wire. The assumed dimensions are L - 3.8 m (length of LE on MIG-31) and radius of a - 0.01 m.
S9
"For these values
(.82(4 = 4 n (3.8)•(251.3 x 0.01)
2
4072 m - 36 db
-
* g
The preceding equation appears in viewgraph 15 of Lecture 2. To estimate width of main lobe, one can use the same formula as used for the inlet. Wires have a lobe structure similar to a flat plate. C
[sin(kL sin 8)
G0
Usin 0
The first zero in RCS occurs when kL sin 0
or where 0 - 0.20 when L - 3.8 m. _
-
The other co-ent is
The second peak occurs at
arcsin(t
The value of the second peak is
u
- 0.28
34 db
that the lobes are very, very narrow.
/1..
*
.
7
/j?7
tLd
*
0 LL
~~~
..
.
..
7 FEATURES OF RC
°..GROSS
FOR FIRE FOX HIG-31
Continuing the calculations, the other leading edges were evaluated as follows: Sweep
a,m 0.02
55300
47
40
10
0.02
450000
56
70
9
0.02
366000
55
90° (wing tip)
1.2
0.01
406
26
330 -,
2
L, m 3.5
40°
,3
o, db
FUSELAGE
The fuselage has a normal vector at an angle of 6 rom pl"3uand 2 - 10 m. So-M
- 94 m2 -20
P
85.
Assume
db
one finds the RCS for fuselage. *
TRAILING EDGE OF PORT WING The TE of port wing is normal to the incident waves from 0 - 1700. Modelling the TE as a wire with L - 12 m and a - 0.01 m, the following results were obtained a a 404000 m 2 . 46 db
"""8 The large RCS is due to high radar frequency. 4-ENGINE EXHAUSTS AT 8 - 1800 The calculation for inlet was repeated with assumed dimensions of 0.8 m x 2.0 m. The result is o
-
2
823550 m2 - 47 db
sm
The RCS have been plotted. The various ak should be added using the formula given in Lecture 2, viewgraph 29. NOTE A note about values of RCS is appropriate.
The leading edges scale as
0 a
a1 Since the wavelength is small,
2
the value of RCS is large.
At X - 1.0 m, the cross sections would be reduced by a factor of (0.025/1.0) 2 - 6.25E-4 or - 32 db. One could
O
subtract 32 db
sm
from kmah va'lue shown for LE or TE if ý were 1.0 m.
~~~~~
-.
'
--
1r7-
~111
I
T-
it
-'---K-
ell
-7
-L
-1-l-
-I
_
~~i44-
LV
44-
-f--
-
-I-I-
T:±. IU
--!
zr++--+t{9____ WZ
1~
7-
II
K
CA
...................................--.... ,.• __ _ ,'''•
•
•
"
'•-.........
..............
.
4....--
8
AINNA SCATTERING
0 Structural Scatterin Term is due to currents iniduced iu the antenna surface 0 and is
'.
0
independent of antenna load impedance.
Antenna Scattering Term is due to current induced at the antenna load terminals.
•.An
antenna launuhes a plane wave from the antenna focus when radiating. power moves outward from focal point to beam. with plane waves,
certain impedance.
The
When radar illuminates the antenna
the power moves to the focus.
The antenna fecd system has a
Depending on the impedance of the feed, the waves may or may
not be re-radiated. !
0
The RCS of an antenna is
given by o -
where a
[
i•]2 ;-+ 8
rd-e expixp
iU RCS of structural scattering term and a
of antenna.
The phase angle between a and a S
e
is the effective ecdo area
is J. The value for a
e
is, for
certain specific conditious,
,2 ^ G2 e
4n
where G is &atenna gain at A. As an estimate, one can use 22 IIT for antenna RCS. 0
As an example, consider an antenna with a gain G of 100 at A the antenna is
eatimated to be 0.) 2100)
IT
4
8 2
0.5 m.
The RCS of
741
Q~"4
IX
ljj
-
Il
A.
E
UHV
S.°
9
.
"RADAR CROSS SECTION REDUCTION -
0
SHAPING implies control of geometry so as tc reduce RCS.
*
0
RAM is material used to maLch rave impedance of free space or to absorb the
EH wave energy. 0
IMEDAUCE LOADING consists o5 passive or active elements added at appropriate locations to control RCS.
0o
S
0
A:E* FUHS
Laj
tJz
ItIA
W
U 4
LIP)
cZZý
uj
QU d4
'ý0
*N.
10 I14PEDAI4CE LOADING 0
On the left-hand side is the case of a loaded pair of cylinders. choice of ZL
By correct
the load impedance, the RCS is reduced by 35 db.
0 On the right-hand side is the case of a pair of circular oglives back-to-back. The incident waves are arriving from the left when 6 - 00. length X/4 is added to the tail end of the body.
A wire with
The wire causes a major
reduction in RCS for 0 < 8 < 45°.
This is an example of scattering by
traveling waves when 0 is small.
A relatively mi-ir change makes a very
large change in RCS. O For simple cases, one may be able to exploit the me~thod. impedance loading to complex shapes may not be obvious.
6q
6'
Application of
'S.j
ct
ui
Iw
N
i
010
N
41
18
-
